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\title{\heiti\zihao{2} 习题11.3 }
\author{中书君}
\date{\songti \today}

\begin{document}
\maketitle
\section{利用$\mathrm{Cauchy}$收敛原理,判断下列级数的敛散性.}
\subsection{ $\sum\limits_{n=1}^{\infty} \frac{\sin n !}{n !}$}
\textbf{解}\quad
$$
\left| \sum\limits_{k=n+1}^{p} \frac{\sin k !}{k !} \right|\leqslant\left| \sum\limits_{k=n+1}^{p} \frac{1}{k !} \right|\leqslant\left| \sum\limits_{k=n+1}^{p} \frac{1}{(k-1)k} \right|<\frac{1}{n}
$$
$\forall\varepsilon>0$取$N=\frac{1}{\varepsilon}+1$,则$\forall n>N$时,$\left|\sum\limits_{k=n+1}^{p}\right|<\varepsilon$,从而级数收敛.


\subsection{ $\sum\limits_{n=1}^{\infty} \frac{\sin n}{2^{n}}$}
\textbf{解}\quad
$$
\left| \sum\limits_{k=n+1}^{p} \frac{\sin k}{2^{k}} \right|\leqslant\left| \sum\limits_{k=n+1}^{p} \frac{1}{2^{k}} \right|<\frac{1}{2^{k-1}}
$$
$\forall\varepsilon>0,\exists N\forall n>N$时,$\left|\sum\limits_{k=n+1}^{p}\right|<\varepsilon$,从而级数收敛.


\subsection{ $\sum\limits_{n=1}^{\infty} \frac{\cos n+\sin n}{n(n+\sin n)}$}
\textbf{解}\quad
$$
\left| \sum\limits_{k=n+1}^{p} \frac{\cos k+\sin k}{n(n+\sin k)} \right|\leqslant\left| \sum\limits_{k=n+1}^{p} \frac{2}{n(n-1)}\right|<\frac{2}{n}
$$
$\forall\varepsilon>0,\exists N\forall n>N$时,$\left|\sum\limits_{k=n+1}^{p}\right|<\varepsilon$,从而级数收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{1}{2 n-1}$}
\textbf{解}\quad
$$
\left| \sum\limits_{k=n+1}^{p}\frac{1}{2 n-1} \right|\leqslant\frac{1}{2}\left| \sum\limits_{k=n}^{p}\frac{1}{k} \right|
$$
当$p\to\infty$时级数发散.

\section{若对任意的 $\varepsilon>0$ 和正整数 $p$,都存在 $N(\varepsilon, p)$,使得当 $n>N(\varepsilon, p)$ 时,总有:$$\left|x_{n+1}+x_{n+2}+\cdots+x_{n+p}\right|<\varepsilon$$成立,问级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 是否收敛.}
\textbf{解}\quad
不一定,这样的$p$依赖于$n$的选取.

\section{设 $x_{n}<z_{n}<y_{n}, n=1,2, \cdots,$ 证明:如果 $\sum\limits_{n=1}^{\infty} x_{n}, \sum\limits_{n=1}^{\infty} y_{n}$ 收敛, 则有 $\sum\limits_{n=1}^{\infty} z_{n}$ 也收敛.}
\textbf{证}$1^{\circ}$\quad
必然有
$$
\left| \sum\limits_{k=n+1}^{p}x_{n} \right|\leqslant\left| \sum\limits_{k=n+1}^{p}z_{n} \right|\leqslant\left| \sum\limits_{k=n+1}^{p}y_{n} \right|
$$
或
$$
\left| \sum\limits_{k=n+1}^{p}x_{n} \right|\geqslant\left| \sum\limits_{k=n+1}^{p}z_{n} \right|\geqslant\left| \sum\limits_{k=n+1}^{p}y_{n} \right|     
$$
由$\mathrm{Cauchy}$收敛原理知其收敛.

\textbf{证}$2^{\circ}$\quad
$$\because x_{n}<z_{n}<y_{n} , \therefore 0<z_{n}-x_{n}<y_{n}-x_{n}$$

由于 $\sum\limits_{n=1}^{\infty} x_{n}$ 和 $\sum\limits_{n=1}^{\infty} y_{n}$ 收敛,故 
$$
\sum\limits_{n=1}^{\infty}\left(y_{n}-x_{n}\right)
$$ 
收敛.由正项级数的比较判别法得 
$$
\sum\limits_{n=1}^{\infty}\left(z_{n}-x_{n}\right)
$$ 
收敛.又因为
$$
\sum\limits_{n=1}^{\infty} z_{n}=\sum\limits_{n=1}^{\infty}\left(z_{n}-x_{n}\right)+\sum\limits_{n=1}^{\infty} x_{n}
$$ 
故$\sum\limits_{n=1}^{\infty} z_{n}$ 收敛

\section{设 $f(x)$ 在 $[-1,1]$ 上具有二阶连续导数, 且 $\lim\limits_{x \rightarrow 0} \frac{f(x)}{x}=0 .$ 证明 : 级数 $\sum\limits_{n=1}^{\infty} f\left(\frac{1}{n}\right)$ 绝对收敛.}
\textbf{证}\quad
$$
f(\frac{1}{n})=f(0)+f'(0)\frac{1}{n}+\frac{f''(0)}{2}\frac{1}{n^{2}}+o\left(\frac{1}{n^{2}}\right)
$$
由于$f(0)=0,f'(0)=\lim\limits_{x\to 0}\frac{f(x)-f(0)}{x-0}=0$,得:
$$
\left|f(\frac{1}{n})\right|=\left|\frac{f''(0)}{2}\frac{1}{n^{2}}+o\left(\frac{1}{n^{2}}\right)\right|\sim \left|\frac{f''(0)}{2}\right|\frac{1}{n^{2}}
$$
由于$\sum\limits_{n=1}^{\infty}\frac{1}{n^2}$收敛,从而级数收敛.

\section{讨论下列级数的敛散性: }
\subsection{ $\sum\limits_{n=1}^{\infty}(-1)^{n-1} \sin \frac{1}{n} $}
\textbf{解}\quad
由于$\sin\frac{1}{n}$单减趋于$0$,从而由$\mathrm{Leibniz}$判别法知级数收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\ln ^{2} n}{n}$}
\textbf{解}\quad
记$f(x)=\frac{\ln^{2}x}{x}$,则$f'(x)=\frac{2\ln x-\ln^{2}x}{x^{2}}$当$x>9$时恒负,从而$n>9$时$\frac{\ln^{2}n}{n}$单减趋于$0$,从而由$\mathrm{Leibniz}$判别法知级数收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}(-1)^{n-1} \frac{\sqrt{n}}{n+1} $}
\textbf{解}\quad
$$
\frac{\sqrt{n}}{n+1}=\frac{1}{\sqrt{n}+\frac{1}{\sqrt{n}}}
$$
由对勾函数性质知其单减趋于$0$,从而由$\mathrm{Leibniz}$判别法知级数收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}(-1)^{n-1} \frac{1}{\sqrt[n]{n}}$}
\textbf{解}\quad
由于$\lim\limits_{n \to \infty}  \frac{1}{\sqrt[n]{n}}\neq 0$,从而级数发散.

\subsection{ $\sum\limits_{n=2}^{\infty} \frac{(-1)^{n-1}}{n \ln n}$}
\textbf{解}\quad
$\frac{1}{n\ln n}$单减趋于$0$,从而由$\mathrm{Leibniz}$判别法知级数收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}(-1)^{n}\left[\mathrm{e}-\left(1+\frac{1}{n}\right)^{n}\right]$}
\textbf{解}\quad
$e-\left(1+\frac{1}{n}\right)^{n}$单减趋于$0$,从而由$\mathrm{Leibniz}$判别法知级数收敛.

\section{设 $\left\{x_{n}\right\}$ 为单调递减非负数列,且有级数 $\sum\limits_{n=1}^{\infty}(-1)^{n-1} x_{n}$ 发散,问级数 $\sum\limits_{n=1}^{\infty}\left(\frac{1}{1+x_{n}}\right)^{n}$ 是否收敛? 请说明理由.}
\textbf{解}\quad
若$x_{n}$有某项为$0$,则后面所有项全部为$0$,与级数$\sum\limits_{n=1}^{\infty}(-1)^{n-1} x_{n}$ 发散矛盾.由单调有界原理知$\left\{x_{n}\right\}$必有唯一聚点.若$\left\{x_{n}\right\}$聚点为$0$,则由$\mathrm{Leibniz}$判别法知级数收敛,矛盾,从而其极限(下确界)为$A>0$.

$$\sum\limits_{n=1}^{\infty}\left(\frac{1}{1+x_{n}}\right)^{n}<\sum\limits_{n=1}^{\infty}\left(\frac{1}{1+A}\right)^{n},A>0$$
$\mathrm{RHS}$收敛,由比较判别法知级数收敛.

\section{$\mathrm{Du Bois-Reymond}$判别法:若级数 $\sum\limits_{n=1}^{\infty}\left(x_{n}-x_{n+1}\right)$ 绝对收敛,级数 $\sum\limits_{n=1}^{\infty} y_{n}$ 收敛,证明:级数 $\sum\limits_{n=1}^{\infty} x_{n} y_{n}$ 收敛.}
\textbf{证}$1^{\circ}$\quad
记$\sum\limits_{k=n}^{m}y_{k}=S_{n,m}$.\par 
Abel变换:
$$
\begin{aligned}
      \left|\sum\limits_{k=n}^{n+p}x_{k}y_{k}\right|&=\left|x_{n+p}S_{n,n+p}+\sum\limits_{k=n}^{n+p-1}(x_{k}-x_{k+1})S_{N,k}-x_{n}S_{N,n-1 }\right|\\
      &\leqslant|x_{n+p}|\cdot|S_{n,n+p}|+\sup_{n\leqslant k\leqslant n+p-1}|S_{N,k}|\sum\limits_{k=n}^{n+p-1}|(x_{k}-x_{k+1})|+|x_{n}|\cdot|S_{N,n-1}|\\
      &\leqslant\sup_{n-1\leqslant k\leqslant n+p}|S_{N,k}|\cdot \left|\left[x_{n}+x_{n+p}+\sum\limits_{k=n}^{n+p-1}|x_{n}-x_{n+1}|\right]\right|
\end{aligned}
$$
由于级数$\sum\limits_{n=1}^{\infty}\left(x_{n}-x_{n+1}\right)$ 绝对收敛,从而$\lim\limits_{n\rightarrow \infty}x_{n}$存在,记为$A$.从而$\forall \varepsilon >0, \exists N_{1},n>N_{1}$时,有$|x_{n}|,|x_{n+p}|,\sum\limits_{k=n}^{n+p-1}\left|x_{n}-x_{n+1}\right|<M$.\par 
由于级数$\sum\limits_{n=1}^{\infty}y_{n}$收敛,从而$\forall\varepsilon>0,\exists N_{2},n>N_{2}$时,能够满足$\left|\sum\limits_{k=n}^{\infty}y_{k}\right|<\frac{\varepsilon}{3M}$
从而有$n>_{max}\left\{N_{1},N_{2}\right\}$
$$
\sup_{n-1\leqslant k\leqslant n+p}|S_{N,k}|\cdot \left|\left[x_{n}+x_{n+p}+\sum\limits_{k=n}^{n+p-1}|x_{n}-x_{n+1}|\right]\right|<\varepsilon
$$
从而由Cauchy收敛准则,级数$\sum\limits_{n=1}^{\infty} x_{n} y_{n}$ 收敛.

\textbf{证}$2^{\circ}$\quad
令 $S_{n}=\sum\limits_{k=1}^{n} y_{k},$ 由  Abel 引理可得 $\sum\limits_{k=1}^{n} x_{k} y_{k}=\sum\limits_{k=1}^{n} S_{k}\left(x_{k}-x_{k+1}\right)+S_{n} x_{n}$\par 
(1) 由于 $\sum\limits_{k=1}^{\infty} y_{k}$ 收敛,设 $\lim\limits_{n \rightarrow \infty} S_{n}=S$;\par 
(2) 由于 $\sum\limits_{k=1}^{\infty}\left(x_{k}-x_{k+1}\right)$ 收敛, 从而 $\left\{x_{k}\right\}$ 收敛,设 $\lim\limits_{n \rightarrow \infty} x_{n}=x $;\par 
(3) 由于 $\sum\limits_{k=1}^{\infty}\left(x_{k}-x_{k+1}\right)$ 绝对收敛,从而 $\sum\limits_{k=1}^{\infty}\left|x_{k}-x_{k+1}\right|$ 收敛,而 $\sum\limits_{k=1}^{\infty}\left|S_{k}\left(x_{k}-x_{k+1}\right)\right|$\par 
是正项级数,因为 $\lim\limits_{n \rightarrow \infty} \frac{\left|S_{n}\left(x_{n}-x_{n+1}\right)\right|}{\left|x_{n}-x_{n+1}\right|}=|S|,$ 所以由比较判别法可知 $\sum\limits_{n=1}^{\infty}\left|S_{n}\left(x_{n}-x_{n+1}\right)\right|$
收敛,从而可知 $\sum\limits_{n=1}^{\infty} S_{n}\left(x_{n}-x_{n+1}\right)$ 收敛.\par 
又因为 $\lim S_{n} x_{n}=x S,$ 所以有 $\sum\limits_{n=1}^{\infty} x_{n} y_{n}=\sum\limits_{n=1}^{\infty} S_{n}\left(x_{n}-x_{n+1}\right)+\lim\limits_{n \rightarrow \infty} S_{n} x_{n}$.\par             
因为级数 $\sum\limits_{k=1}^{\infty} S_{n}\left(x_{k}-x_{k+1}\right)$ 收敛,数列 $\left\{S_{n} x_{n}\right\}$ 收敛,所以 $\sum\limits_{n=1}^{\infty} x_{n} y_{n}$ 收敛.

\section{讨论下列级数的敛散性}
\subsection{ $\sum\limits_{n=2}^{\infty} \frac{\sin \frac{n \pi}{4}}{\ln n}$}
\textbf{解}\quad
级数$\sum\limits_{n=1}^{\infty}\sin \frac{n\pi}{4}$乘以$\sin\frac{\pi}{8}$即可化为裂项形式,即公差的$\frac{1}{2}$.以后默认这种级数有界.而对于$\frac{1}{\ln n}$单调趋于$0$,从而由$\mathrm{Dirichlet}$判别法知级数收敛.

而又因为
$$
\left|\frac{\sin \frac{\pi}{4}}{\ln n}\right|>\left|\frac{\sin^{2} \frac{\pi}{4}}{\ln n}\right|=\frac{1-\cos\frac{n\pi}{2}}{2\ln n}=\frac{1}{2\ln n}-\frac{\cos \frac{n\pi}{2}}{2\ln n}
$$前者发散后者收敛,从而有级数条件收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\cos 2 n}{n^{p}} $}
\textbf{解}\quad
级数$\sum\limits_{n=1}^{\infty}(-1)^{n}\cos 2n$有界,$\frac{1}{n^{p}}$单调趋于$0$,(将其视为$\cos 2n+ n\pi$)由$\mathrm{Dirichlet}$判别法知级数收敛.

显然当$p>1$时,可知原级数绝对收敛(比较判别法),当$0<p\leqslant 1$时,原级数条件收敛(证明方法同8.1),当$p\leqslant 0$时,$\lim\limits_{n\to \infty}(-1)^{n}\frac{\cos 2n}{n^{p}}\neq 0$,级数发散.

\subsection{ $\sum\limits_{n=2}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) \frac{\sin n x}{n} $}
\textbf{解}\quad
$a_{n}=\sin nx$的级数有界.对于$\frac{\sum\limits_{k=1}^{n}\frac{1}{k}}{n}$,显然将其视为$\sum\limits_{k=1}^{n}\frac{1}{k}$的平均值在单调减少,由$\mathrm{Stolz}$定理知其极限为$0$,从而由$\mathrm{Dirichlet}$判别法知级数收敛.

对此类带三角函数的级数同上方法可知其条件收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\cos ^{2} n}{n^{p}} $}
\textbf{解}\quad
当$p\leqslant 0$时,由于$\lim\limits_{n \to \infty} (-1)^{n} \frac{\cos ^{2} n}{n^{p}} \neq 0$,级数发散.

当$0<p\leqslant 1$时,由于$ (-1)^{n} \frac{\cos ^{2} n}{n^{p}} =\frac{\cos 2n}{2n^{p}}+\frac{(-1)^{n}}{2n^{p}}$,前者显然由$\mathrm{Leibniz}$判别法可知其级数条件收敛.而对于级数$$\sum\limits_{n=1}^{\infty}\left|\frac{\cos^2 n}{n^{p}}\right|=\sum\limits_{n=1}^{\infty}\frac{\cos^2 n}{n^{p}}=\sum\limits_{n=1}^{\infty}\frac{1}{2n^{p}}+\frac{\cos 2n}{2n^{p}}$$前者发散,后者收敛.所以级数条件收敛.

当$p>1$时,同上,级数收敛.而此时$\sum\limits_{n=1}^{\infty}\frac{1}{2n^{p}}$绝对收敛,从而易知级数绝对收敛.

\subsection{ $\sum\limits_{n=2}^{\infty}\left(\frac{\sin n x}{\sqrt{n}}+\frac{\sin ^{2} n x}{n}\right)$}
\textbf{解}\quad
由$\mathrm{Dirichlet}$判别法易知
$$
\sum\limits_{n=2}^{\infty}\frac{\sin n x}{\sqrt{n}}
$$
收敛,又由于
$$
\sum\limits_{n=2}^{\infty}\frac{\sin ^{2} n x}{n}=\sum\limits_{n=2}^{\infty}\frac{1-\cos 2 n x}{2n}=\sum\limits_{n=2}^{\infty}\frac{1}{2n}-\sum\limits_{n=2}^{\infty}\frac{\cos 2nx}{2n}
$$
前者发散,后者收敛,从而级数发散.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{\sin (n+1) x \cos (n-1) x}{n^{p}}$}
\textbf{解}\quad
$$\sin (n+1)x\cos (n-1)x=\frac{1}{2}[\sin(2nx)+\sin 2x]$$
当$p\leqslant 0$时,$\lim\limits_{n \to \infty} \frac{\sin (n+1) x \cos (n-1) x}{n^{p}}\neq 0$从而级数发散.

当$p>1$时,由比较判别法知级数绝对收敛.

当$0<p\leqslant 1$时,若$\sin 2x\neq 0$,则$\sum\limits_{n=1}^{infty}\frac{1}{n^{p}}$发散,$\sum\limits_{n=1}^{\infty}\frac{\sin 2nx}{n^{p}}$收敛,从而级数发散.而若$\sin 2x = 0$,则$\sin 2nx=0$,级数为$0$,收敛.

\section{已知级数 $\sum\limits_{n=1}^{\infty} \frac{x_{n}}{n^{b}}$ 收敛,证明 : 当 $a>b$ 时,级数 $\sum\limits_{n=1}^{\infty} \frac{x_{n}}{n^{a}}$ 也收敛.}
\textbf{证}\quad
显然$\frac{1}{n^{a-b}}$单减趋于$0$,从而由$\mathrm{Abel}$判别法知$$\sum\limits_{n=1}^{\infty} \frac{x_{n}}{n^{b}}\cdot \frac{1}{n^{a-b}}=\sum\limits_{n=1}^{\infty} \frac{x_{n}}{n^{a}}$$收敛.


\section{使用 $\mathrm{Abel}$ 引理证明 $\mathrm{Abel}$ 判别法.}
一般我们说Abel变换较多,因为许多东西都是其直接衍生物.

\textbf{证}\quad
设$\sum\limits_{k=1}^{\infty}x_{k}$收敛,极限为$X$,$y_{n}$单调有界,极限为$Y$.由于$\sum\limits_{k=1}^{\infty}x_{k}y_{k}=XY+\sum\limits_{k=1}^{\infty}S_{k}(y_{k}-y_{k+1})$,只需证$\sum\limits_{k=1}^{\infty}S_{k}(b_{k}-b_{k+1})$收敛.

取得到$N_{1}$,当$n>N_{1}$时,$S_{n}<S+1$.

$$
\forall\varepsilon>0,\exists N_{2},\mathrm{s.t.}\forall n>N_{2},|y_{n}-Y|<\frac{\varepsilon}{2(S+1)}
$$
从而$|y_{n}-y_{n+1}|<\frac{\varepsilon}{S+1}$.

由$\{y_{n}\}$单调性(不妨单减)可知
$$
n>_{max}\{N_{1},N_{2}\},\sum\limits_{k=n+1}^{\infty}S_{k}(y_{k}-y_{k+1})<\varepsilon
$$
从而级数收敛.证毕.

\section{已知级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 发散,证明 :级数 $\sum\limits_{n=1}^{\infty}\left(1+\frac{1}{n}\right) x_{n}$ 也发散.}
\textbf{反证}\quad
可由题$7\quad\mathrm{Du Bois-Reymond}$判别法:若$\sum\limits_{n=1}^{\infty}\left(1+\frac{1}{n}\right) x_{n}$收敛,取$b_{n}=\frac{1}{1+\frac{1}{n}}=\frac{n}{n+1}$,则由于
$$
\sum\limits_{n=1}^{\infty}b_{n+1}-b_{n}=\sum\limits_{n=1}^{\infty}\frac{1}{n+1}-\frac{1}{n+2}
$$
绝对收敛可知$$\sum\limits_{n=1}^{\infty}b_{n}\left(1+\frac{1}{n}\right) x_{n}=\sum\limits_{n=1}^{\infty}x_{n}$$收敛,矛盾.证毕.

\section{讨论下列级数的敛散性}
\subsection{ $\sum\limits_{n=2}^{\infty} \frac{\cos 3 n}{n}\left(1+\frac{1}{n}\right)^{n}$}
\textbf{解}\quad
由于 $\sum\limits_{n=2}^{\infty} \cos 3 n$ 有界, $\left\{\frac{1}{n}\right\}$ 单调递减趋于$0$,由狄利克雷判别法可知, $\sum\limits_{n=2}^{\infty} \frac{\cos 3 n}{n}$ 收敛.

又因为 $\left(1+\frac{1}{n}\right)^{n}$ 单调有界,又由阿贝尔判别法可知,原级数收敛. $$\because \left|\frac{\cos 3 n}{n} \cdot\left(1+\frac{1}{n}\right)^{n} \right| >\frac{2 \cos ^{2} 3 n}{n}=\frac{1+\cos 6 n}{n}, \sum\limits_{n=1}^{\infty} \frac{1}{n}$$ 发散, 而 $$\sum\limits_{n=2}^{\infty} \frac{\cos 6 n}{n}$$ 收敛$\therefore$ 原级数条件收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\cos ^{2} n}{n} \arctan (3+n)$}
\textbf{解}\quad
由于
$$
\begin{aligned}
    (-1)^{n} \frac{\cos ^{2} n}{n} \arctan (3+n)&=(-1)^{n} \frac{\cos 2 n+1}{2 n} \arctan (3+n)\\
    &=(-1)^{n} \frac{1}{2 n} \arctan (3+n)+(-1)^{n} \frac{\cos 2 n}{2 n} \arctan (3+n)
\end{aligned}
$$
又因为
$$\left\{\frac{1}{2 n}\right\}$$ 递减趋于$0$,故由$\mathrm{Leibniz}$判别法 $$\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{1}{2 n}$$ 收敛,而 $\{\arctan (3+n)\}$ 单调有界, 则 $$\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{1}{2 n} \arctan (3+n)$$ 收敛.

同理,易知 $$\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\cos 2 n}{2 n} \arctan (3+n)$$收敛.
则有
$$ \sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\cos ^{2} n}{n} \arctan (3+n)$$ 
收敛.
$$
\because\left|(-1)^{n} \frac{\cos ^{2} n}{n} \arctan (3+n)\right|=\frac{\cos 2 n+1}{2 n} \arctan (3+n)
$$ 
其中 
$$
\sum\limits_{n=1}^{\infty} \frac{\cos 2 n}{2 n} \arctan (3+n)
$$ 
收敛.

$0<\frac{1}{2 n} \arctan (3+n)<\frac{1}{n}$ ,而 $\sum\limits_{n=1}^{\infty} \frac{1}{n}$发散, 则 $\sum\limits_{n=1}^{\infty} \frac{1}{2 n} \arctan (3+n)$ 发散,从而原级数条件收敛.

\subsection{ $\sum\limits_{n=2}^{\infty}(-1)^{n} \frac{\sin ^{2} n}{\ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)$}
\textbf{解}\quad
$$
\begin{aligned}
    (-1)^{n} \frac{\sin ^{2} n}{\ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)=&(-1)^{n} \frac{1-\cos 2 n}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)\\
    =&\frac{(-1)^{n}}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)\\
    &-(-1)^{n} \frac{\cos 2 n}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)\\
    =&\frac{(-1)^{n}}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)\\
    &-\frac{\cos (2 n+n \pi)}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)
\end{aligned}
$$
而 
$$
\sum\limits_{n=1}^{\infty}(-1)^{n}, \sum\limits_{n=1}^{\infty} \cos (2 n+n \pi)
$$ 
有界, $\left\{\frac{1}{2 \ln n}\right\}$ 单调递减趋于$0$,由$\mathrm{Dirichlet}$判别法得
$$
\sum\limits_{n=1}^{\infty} \frac{(-1)^{n}}{2 \ln n}, \sum\limits_{n=1}^{\infty} \frac{\cos (2 n+n \pi)}{2 \ln n}
$$ 
收敛.因为 
$$
\left(1+\frac{1}{n}\right)^{n}, 3-\arctan n
$$ 单调有界,由阿贝尔判别法 
$$
\frac{(-1)^{n}}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n), \frac{\cos (2 n+n \pi)}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)
$$ 
收敛.
由此可得 
$$ \sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\sin ^{2} n}{\ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)$$ 
收敛.又由于
$$
\left|(-1)^{n} \frac{\sin ^{2} n}{\ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)\right|=\frac{1-\cos 2 n}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)
$$
其中 $\frac{\cos 2 n}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)$ 收敛,
$$
\sum\limits_{n=1}^{\infty} \frac{1}{2 \ln n}\left(1+\frac{1}{n}\right)^{n}(3-\arctan n)
$$ 
发散,从而原级数条件收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\sin ^{2} 3n}{n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n,$ 其中 $\alpha>0, p>0$}
\textbf{解}\quad
当 $0 < p \leqslant 1$时,有
$$
\begin{aligned}
    \sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\sin ^{2} 3n}{n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n=&\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{1-\cos 10 n}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n\\
    =&\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{1}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n\\
    &-\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{\cos 10 n}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n\\
    =&\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{1}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n\\
    &-\sum\limits_{n=1}^{\infty} \frac{\cos (10 n+n\pi)}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n
\end{aligned}
$$
由于
$$
\sum\limits_{n=1}^{\infty}(-1)^{n},\sum\limits_{n=1}^{\infty}\cos (10n+n\pi)
$$
有界,$\frac{1}{2n^{p}+\alpha}$单减趋于$0$,从而由$\mathrm{Dirichlet}$判别法知级数
$$
\sum\limits_{n=1}^{\infty}\frac{(-1)^{n}}{2n^{p}+\alpha},\sum\limits_{n=1}^{\infty}\frac{\cos (10n+n\pi)}{2n^{p}+\alpha}
$$
收敛.又因为
$$
\left(1+\frac{1}{n}\right)^{n},\arctan 3n
$$
单调有界,从而由$\mathrm{Abel}$判别法知级数
$$
\sum\limits_{n=1}^{\infty}(-1)^{n} \frac{1}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n,\sum\limits_{n=1}^{\infty} \frac{\cos (10 n+n\pi)}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n
$$
收敛.又因为
$$
\sum\limits_{n=1}^{\infty}\left|(-1)^{n} \frac{\sin ^{2} 3n}{n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n\right|=\sum\limits_{n=1}^{\infty}\frac{1-\cos 10 n}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n
$$
其中
$$
\sum\limits_{n=1}^{\infty}\frac{1}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n
$$
发散,
$$
\sum\limits_{n=1}^{\infty}\frac{\cos 10 n}{2n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3 n
$$
收敛,从而级数条件收敛.

当 $p>1$时,
$$\left|(-1)^{n} \frac{\sin ^{2} 3n}{n^{p}+\alpha}\left(1+\frac{1}{n}\right)^{n} \arctan 3n \right|=\frac{1-\cos 10 n}{2\left(n^{p}+\alpha\right)}\left(1+\frac{1}{n}\right)^{n} \arctan 3n$$

由于 
$$
\frac{1}{2\left(n^{p}+\alpha\right)}<\frac{1}{n^{p}}, \sum\limits_{n=1}^{\infty} \frac{1}{n^{p}}
$$ 
收敛, 由比较判别法,正项级数 $\sum\limits_{n=1}^{\infty} \frac{1}{2\left(n^{p}+\alpha\right)}$收敛而 $\left(1+\frac{1}{n}\right)^{n}$ 单调有界,由$\mathrm{Abel}$判别法, 
$$ 
\sum\limits_{n=1}^{\infty} \frac{1}{2\left(n^{p}+\alpha\right)}\left(1+\frac{1}{n}\right)^{n}
$$ 
收敛.同理 
$$
\sum\limits_{n=1}^{\infty} \frac{1}{2\left(n^{p}+\alpha\right)}\left(1+\frac{1}{n}\right)^{n} \arctan 3n
$$ 
收敛.类似可证 
$$
\sum\limits_{n=1}^{\infty} \frac{\cos 10 n}{2\left(n^{p}+\alpha\right)}\left(1+\frac{1}{n}\right)^{n} \arctan 3n
$$ 
收敛, 从而原级数绝对收敛.

\end{document}